Arrange the following carbanions in the order | vedclass

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(C) The stability of a carbanion is directly proportional to the $s$-character of the carbon atom bearing the negative charge.$i$. $RC \equiv C^{\omin

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$i > ii > iii > iv$

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(C) The stability of a carbanion is directly proportional to the $s$-character of the carbon atom bearing the negative charge.$i$. $RC \equiv C^{\ominus}$: The carbon is $sp$ hybridized ($50\% \ s$-character).$ii$. $C_6H_5^{\ominus}$: The carbon is $sp^2$ hybridized ($33.3\% \ s$-character).$iii$. $R_2C = CH^{\ominus}$: The carbon is $sp^2$ hybridized ($33.3\% \ s$-character). However,the alkyl groups $(R)$ are electron-donating,which destabilizes the carbanion compared to $C_6H_5^{\ominus}$.$iv$. $R_3C - CH_2^{\ominus}$: The carbon is $sp^3$ hybridized ($25\% \ s$-character),which is the least stable.Therefore,the correct order of stability is $i > ii > iii > iv$.

Arrange the following carbanions in the order of their stability:
$i$. $RC \equiv C^{\ominus}$
$ii$. $C_6H_5^{\ominus}$
$iii$. $R_2C = CH^{\ominus}$
$iv$. $R_3C - CH_2^{\ominus}$

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Arrange the following carbanions in the order of their stability:$i$. $RC \equiv C^{\ominus}$$ii$. $C_6H_5^{\ominus}$$iii$. $R_2C = CH^{\ominus}$$iv$. $R_3C - CH_2^{\ominus}$